1 and 4
8m
 Area is LxW
 So working backwards, 72=9xW
 Rearrange. W=8
72 stickers
 80/10 = 8
 45/5 = 9
 8 * 9 = 72
A
36cm² as area means sides must be 6cm long each.
Therefore, total is 24cm
B
 180 – 25 – 107 = 48
44 stickers
 Use division as a starting point
 So 60/15 = 4
 55/5 = 11
 Multiply the 2 to get 44 stickers
c) 55
Notice the pattern and sequence. 1 gives 4 gives 9 gives 16. So a change of 3, 5, 7, 9 etc. Apply this pattern
a) See below
b) See below
c) 40
10 x 4
d) 5,050
This can be done with the following calculation
(100×101)/ 2 = 101 x 50 = 5,050
64cm³
The volume of Cube A is 8cm³. So this is 2x2x2. For 64cm, think of cube numbers. So 2x2x2 = 8. 3x3x3 = 27. 4x4x4=64
So double the side of Cube A is 4 so its 64cm³
a) 48cm
See labelling below
b) 108cm²
Area = WxL. 6×18
c) 18
In 18cm across, she can fit 6 (18/3). Given its 6cm tall, she can fit 3 high. So in total 3 rows each with 6 in each row 18
a) See below
Its 2 4×7 triangles that have been transposed from the original question
b) See below
a) See below
b) (2,7)
c) See above for drawing
A pentagon can have 5 obtuse angles but not always has to be the case
25cm²
If perimeter is 20cm, each side is 5cm
Therefore, the area is 5 x 5 = 25cm²
5/8ths is shaded
6
A is 6 x 2. Given we have a 3 x 1, we can fit in 4 rectangles into this side
B is 3 x 2. Given we have a 3 x 1, we ca fit in 2 rectangles into this side
a) 9cm
Perimeter of 36cm . 4 sides. So 9cm per side (36/4)
b) 81cm²
9×9
a) 32cm²
Area of a rectangle = LxW
b) 4cm²
8 triangles altogether = 1/2 LxW
c) 8cm²
4 and 5 in the above
d) 12
6 small triangles from 1,2,3,6,7,8
4 medium triangles from (1,2) , (2,3), (6,7), (7,8)
2 larger triangles from (1,4,5,6) and (3,4,5,8)
a) 68cm²
 Firstly we need to count the number of cubes. This is equal to 30
 For the 30 cubes, these can be split as follows:

 There are 12 cubes….the steps if you look from South East to North West angle. 2 sides visible so thats 24 in total
 Next the cubes at the side looking from South West to North East angle. There are 6 visible cubes and in total, 10 surfaces. This is the same on the other side so its 20 visible sides in total
 Finally, 6 cubes have 4 sides showing
 Total therefore is 68 sides showing

b) 3 cubes left
Easier than part a for sure!. 30 cubes in part a and 27 in part b
a) 8 small boxes in the bottom layer
 Given the height of the cube is 2cm, a total of 3 high can fit in the larger box
 The bottom layer of the larger box would therefore occupy 2 x 5 x 9 = 90cm³
 But some space would be left around the base. Below is an aerial view
b) 24 small boxes
 As mentioned above, 3 layers in total
 Therefore 8 x 3 =24
 A 1 x 1 would be left as empty space as you would not be able to break a small box and fit it into the sides
a) A & C. Both have a perimeter of 8
b) 4cm²
a) 87m²
 The area of the lawn is 96m² (12 x 8)
 The patio will take 3 x 3 = 9m²
 Therefore, the remaining area of the lawn is 96 – 9 = 87m²
b) 40m
 The perimeter of the lawn is 2 x (12 + 8) = 40m
a) 25%
 Firstly count the number of segments. There are 12 segments
 3 of the segments are shaded
 Therefore in total, 3/12 are shaded = 25%
b) 3/16
 Each Large Triangle is equivalent to 2 small triangles
 There are 3 small triangles shaded out of a total of 16
c) See below
 16 squares altogether
 4/16 = 25% shaded
a) 10cm
 This is a classic question and worth remembering the process to tackle these.
 Perimeter is based on 16 sides.
 Given perimeter is 160cm, each side is 160/16 = 10cm in length
b) 100cm²
10×10
c) 1600cm²
If each Square is 100cm², area of whole figure is 16 x 100
a) x = 2cm, y = 6cm
x = 2cm (8cm – A – B)
y = 6cm (14cm – E)/2
b) 52cm
Perimeter starting at point A = 2+6+2+2+2+6+4+14+4+6+2+2
c) 72cm²
Area = (14×4) + (2×2) + (6×2) = 56 + 4 + 12
16units²
 Break it down into 2 parts
 Shape A = 1/2 x Width x Height. So this is 1/2 x 4 x 4 = 8
 Shape B = 2 x 4 = 8
5 cubes
 Tall Cube. Number of cubes is W * L * H. This equals 2 * 2 * 8. = 32
 Small Cube. Number of cubes is W * L * H. This equals 3 * 3 * 3 = 27
20cm
 Very Standard style question where you are given the area and have to work out the perimeter.
 x * 4 = 24cm² would give the area. So rearrange this. Where x = Length. 4 is Width
 24/4 = x
 x = 6
 So if x = 6, the solution is [(2 * 6) + (2 * 4) ]
 Area of the shape is 18 full squares + 7 other squares from partially filled squares
 Total area = 25 squares
a) 15/23
 Consider total area of shape = 23 squares
 Shaded area is equal on all 3 sides. Each side has 5 shaded squares
 Total shaded area = 15/23
b) See below
 16 triangles in total
 3/4 shaded = 12/16 shaded
c) i) Shape A DOES NOT have a greater fraction shaded than Shape B
 Shape A – 10 squares altogether. 2/10 shaded = 1/5th shaded
 Shape B – 16 squares altogether. 4/16 shaded = 1/4 shaded
c) ii) Shape B DOES have a greater fraction shaded than Shape A. (Inverse of the statement for i). So this is the right statement
c) iii) This is NOT CORRECT
a) 14
A diagonal is a line that connects two nonadjacent corners. All polygons, except for the triangle, have a number of diagonals. The formula to find the number of diagonals is n(n – 3)/2, where n is the number of sides the polygon has
b) 54
So using the above formula, 12(123)/2 = (14436)/2 = 108/2
a) 21cm
6 x 3.5cm = 21cm
b) 63cm
Pattern 2 has 3 sides per hexagon on the outside. So its 18 sides altogether. 18 X 3.5
c) 19 hexagons
Pattern 3 will require 12 additional hexagons. 1 hexagon will cover 2 of the outsides and the next one 1 outside of pattern 2. And so on. So it will be 1+6+12 (Pattern 1 + Pattern 2 + Pattern 3)
d) 105cm
30 sides will be showing for Pattern 3 on the outside. So 30 * 3.5
8cm
If Square area is 36cm², then each side is 6cm. Perimeter is therefore 24cm. Given the perimeter of the triangle is the same, each side of the triangle must be 24/3 = 8cm
a) See the diagram below
b) (5,4)
c) 17.5cm²
Area: Point A (1,5); B (4,9); C (8,8); D (5,4)
2 Triangles in effect with ABD & BCD.
ABD Area – A to midpoint BD is 3.5. Base is 5 for B to D (coordinate for Y axis for B of 9 less coordinate for D of 4)
So area of ABD = (1/2 * 3.5 * 5) = 15.75 * 2 = 8.75. * 2 as there are 2 triangles = 17.5
26cm²
Area covered in paint if all 6 cubes are dipped in paint is W*L*D = (1*1*1) * 6 sides * 6 cubes = 36
However, there are 10 covered faces. (remember to count those back to back faces as 2). So in total, 10cm²
a) 12
Area of a cube is 3 x 2 x 2 = 12 (L x H x D)
b) 84
6 x 4 x 4 = 96. Less what was used by Tilly = 84
a) 8cm²
Count the number of full squares first = 5
Then pair up the smaller squares to whole ones. The ones on the left come to 2 squares. The ones on the right that are not complete come to 1 square. Total is 8
b) See below
Note the triangle can be many different permutations. One example below
a) octagon
b) obtuse
(greater than 90 degrees)
c) i) 3.4cm
c ii) 27.2cm
34mm x 8 = 272mm
a) 80cm
Quite a tricky question so key to label like for like sides
X = 13cm. There are 4 of these so thats 13 x 4 = 52cm
Y = 125cm (Take away 5 because the height is the central part that is not part of the perimeter. So thats 7 * 4 = 28cm
b) 30cm²
Just divide the total by 4
a) Parallelogram
…A rectangle pushed over sideways
b) 0
A parallelogram has zero lines of symmetry
…A square however pushed over sideways, has 2 lines of symmetry (A rhombus)
c) (2,0)
Always read X Axis first
d) See below
One example of many are below with coordinates below of (8,4). Provided you do not use coordinates of (5,5) for point E, then any replacement of the X Axis will suffice. So eg (1,4), (2,4) etc all work as does (9,4), (10,4) etc
a) 12cm²
Count whole squares first = 11
Partial half squares = 2 halves = 1
b) See below
a) b) c) See below
a) 12cm²
Each shaded box is 1cm x 1cm.
b) 18cm
Ensure you label each side to confirm all sides are included in the calculation
c) 6×2 rectangle drawn
These types of questions are good practice for Non Verbal Reasoning and particularly cube related shape questions which are common.
Direction you need to turn: Anticlockwise
Angle of turn required: 60 degrees
Like prior word based questions, just break it down 1 sentence at a time:
Start at zero degrees (or think of it as 12 o’clock). 40 degrees clockwise will be to the right
70 degrees anticlockwise will take you a net 30 degrees to the left of zero
90 degrees clockwise turn from there will take you to a net 60 degrees right. So 60 degrees clockwise